What is the solution for (x+1)^2y^'+5(x+1)y=16 ?

The solution for (x+1)^2y^'+5(x+1)y=16 is y= 4/(x+1)+(c_{1})/(x^5+5x^4+10x^3+10x^2+5x+1)

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(x+1)^2y^'+5(x+1)y=16

(x + 1)^{2} y^{^{'}} + 5 (x + 1) y = 16

y = \frac{4}{x + 1} + \frac{c _{1}}{x ^{5} + 5 x ^{4} + 10 x ^{3} + 10 x ^{2} + 5 x + 1}

Solution steps

(x + 1)^{2} y^{'} (x) + 5 (x + 1) y = 16

Solve linear ODE:

y = \frac{4}{x + 1} + \frac{c _{1}}{x ^{5} + 5 x ^{4} + 10 x ^{3} + 10 x ^{2} + 5 x + 1}

y = \frac{4}{x + 1} + \frac{c _{1}}{x ^{5} + 5 x ^{4} + 10 x ^{3} + 10 x ^{2} + 5 x + 1}

i = 1 \sum \infty \frac{1}{i ^{2}}

\int e^{- 2 a x} d x

\int (\frac{x ^{2} + 2 x - 3}{x ^{4}}) d x

x = e^{y}, y = 0, x = 1, x = e^{2}

\int e^{- 2 x^{2}} x d x

What is the solution for (x+1)^2y^'+5(x+1)y=16 ?
The solution for (x+1)^2y^'+5(x+1)y=16 is y= 4/(x+1)+(c_{1})/(x^5+5x^4+10x^3+10x^2+5x+1)